lf `f(x) ={ax^2+b ,x leq 1 and bx^2+ax+c , x >1; b != 0`, then `f(x)` is continuous and differentiable at `x = 1` if

Options

A. `c=0,a=2b`
B. `a=b,c in R`
C. `a=b,c=0`
D. `a=b,c ne 0`

Answer

A

Explanation

Since , f(x) is continuous at x = 1
`therefore underset(xto1^(-))lim f(x) = underset(xto1^(+))limf(x)`
`rArr a + b = b + a + c rArr c = 0 `
Here , ` f ' (x) = {{:(2ax",",a ne ","x lt 1),(2bx + a",",xgt 1):}`
Also , f(x) is differentiable at x = 1
` therefore ` (LHD at x = 1 ) = (RHD at = 1)
`rArr 2a = 2b (1) + a rArr a = 2b ` .